Question
Find the value of:
\[ \tan^{-1}(\tan \tfrac{9\pi}{8}) \]
Solution
The principal value range of \( \tan^{-1}x \) is:
\[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \]
Now reduce the angle:
\[ \frac{9\pi}{8} = \pi + \frac{\pi}{8} \Rightarrow \tan \tfrac{9\pi}{8} = \tan \tfrac{\pi}{8} \]
But \( \tfrac{\pi}{8} \in \left(0, \frac{\pi}{2}\right) \), so
\[ \tan^{-1}(\tan \tfrac{\pi}{8}) = \tfrac{\pi}{8} \]
Final Answer:
\[ \boxed{\tfrac{\pi}{8}} \]
Key Concept
Reduce the angle and then apply principal value range of inverse tangent.